Given number of balls = 3 + 5 + 7 = 15
One ball is drawn randomly = 15C1
probability that it is either pink or red =
Let X be the event that cards are in a club which is not king and other is the king of club.
Let Y be the event that one is any club card and other is a non-club king.
Hence, required probability:
=P(A)+P(B)
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= = =
Let A be the event that X is selected and B is the event that Y is selected.
P(A) = 1/7, P(B) = 2/9.
Let C be the event that both are selected.
P(C) = P(A) × P(B) as A and B are independent events:
= (1/7) × (2/9) = 2/63
Let A, B, C be the events of getting a white ball in first, second and third draw respectively, then
Required probability =
=
Now, P(A) = Probability of drawing a white ball in first draw = 4/18 = 2/9
When a white ball is drawn in the first draw there are 17 balls left in the urn, out of which 3 are white
Since the ball drawn is not replaced, therefore after drawing a white ball in the second draw there are 16 balls left in the urn, out of which 2 are white.
Hence the required probability =
Required probability is given by P(E) =
The possible outcomes are as follows :
5H, 5T, (H, 4T), (T, 4H), (2H, 3T) (3H, 2T), i.e. 6 outcomes in all.
Therefore the probability that head appears an odd number of times = 3/6 =1/2 (In only three outcomes out of the six outcomes, head appears an odd number of times).
The probability of an impossible event is 0.
The event is known ahead of time to be not possible, therefore by definition in mathematics, the probability is defined to be 0 which means it can never happen.
The probability of a certain event is 1.
Find the number of cases in which none of the digits show a '6'.
i.e. all three dice show a number other than '6', 5×5×5=125 cases.
Total possible outcomes when three dice are thrown = 216.
The number of outcomes in which at least one die shows a '6' = Total possible outcomes when three dice are thrown - Number of outcomes in which none of them show '6'.
=216?125=91
The required probability = 91/256
P( only one of them will be selected) = p[(E and not F) or (F and not E)]
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